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Let $n>1$ be an integer, and $a>1$ a real number. Consider the subspace $L<R^{2^n}$ generated by the $n$ possible tensor products of the $n-1$ copies of the vector $(1,a)$ and one copy of $(a,-1)$. (The $n$ generators of $L$ correspond to the $n$ positions where the factor $(a,-1)$ can be inserted into the product. Say, if $n=2$, then $L$ is generated by the two vectors $(a,-1,a^2,-a)$ and $(a,a^2,-1,-a)$.) Now for a vector $l\in L$ and a positive integer $k\le 2^n$, choose arbitrarily $k$ coordinates of $l$ and let $\sigma$ denote their sum. Since $\sigma$ is the scalar product of $l$ and a vector of norm $\sqrt k$, we have $$ |\sigma| \le \|l\| \sqrt k. $$

My question is whether this trivial estimate can be improved by a growing factor; say,

Is it true that for any $l\in L$ and $k\le 2^n$, the sum of any $k$ coordinates of $l$ is at most $C_a\|l\| \sqrt k / \log\log n$ in absolute value, with a constant $C_a$ depending only on $a$?

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    $\begingroup$ What are the coordinates of a vector in $R^{2^n}$ (that should be defined uniformly in $n$, for your question to make sense), do you consider it as the $n$-th tensor power of $R^2$ and generated by the elements $e_{i_1}\otimes e_{i_2}\otimes \dots \otimes e_{i_n}$? $\endgroup$
    – Maurizio Monge
    Commented Apr 15, 2011 at 12:03
  • $\begingroup$ @Maurizio: the vectors, generating $L$, are vectors of the form $e_1\otimes\dots\otimes e_n$, where $n-1$ of the vectors $e_i$ are equal to $(1,a)$, and one of these vectors is equal to $(a,-1)$. So, all vectors $e_i$ lie in $R^2$, and their product lies in $R^{2^n}$. $\endgroup$
    – Seva
    Commented Apr 15, 2011 at 14:28

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I don't think so. Each generator $l$ of $L$ is a vector of length $2^n$ with $\binom{n}{i}$ entries equal to $\pm a^i$ for $0\le i \le n.$ So $\|l\|=(a^2+1)^{n/2}$ and there is only one entry which is $a^n$. If $a$ is large enough, then wouldn't that $k=1$ entry be larger than $\|l\| \sqrt 1 / \log\log n ? $

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  • $\begingroup$ Correct - but, in fact, I didn't mean uniformity in $a$ here. I have inserted a minor, but important refinement. $\endgroup$
    – Seva
    Commented Apr 17, 2011 at 6:17

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